# Central Limit Theorem exercise

I'm trying to solve this exercise:

Drums labeled 30 L are filled with a solution from a large vat. The amount of solution put into each drum is random with mean 30.01 L and standard deviation 0.1 L.

A) What is the probability that the total amount of solution contained in 50 drums is more than 1500 L?

B) If total amount of solution is the vat is 2401 L,what is the probability that 80 drums can be filled without running out?

I have managed to solve A) by applying the Central Limit Theorem:

n = 50

S = "Total amount of solution in 50 drums"

The total amount of solution in n will have a mean of n x mean = 50 x 30.01, and a variance = 50 x 0.01 = 0.5 Using this values I'm able to calculate P(S > 1500) by approximating to a Standard Normal Distribution and checking on the cumulative table, obtaining a result of 0.758.

My problem is on exercise B). Since they ask the opposite, I'm having trouble at figuring out how to start. Does it have anything to do with a binomial distribution?

I'd really appreciate some help :)

For Part A, you took the square root of $n \sigma^2$ before you divided $S - 1500.5$ by it, correct? Under that assumption, what you did looks good. (I got a slightly different answer, approximately $0.760$ — hence my question. I'm guessing it's round-off, though.)
Your observation for Part B is spot-on, perhaps more so than you realize :). Let $S_{80}$ be the total number of liters in $80$ barrels. If only $2401$ liters are available to fill those barrels, the amount that will go into those barrels cannot exceed $2401$; in other words, we want $P(S_{80} \leq 2401)$. You'd use the same setup, namely, the probability that
$$\frac{S_{n} - n \mu}{\sqrt{n} \sigma } \leq \frac{2401 - n \mu}{\sqrt{n} \sigma}$$
• I probably won't see your response till much later today, but I got approximately $0.588$, which is a bit more than round-off error. Are you calculating $Z \leq (2401 - 80\cdot 30.01) / (\sqrt{80} \cdot 0.1)$? – dmk Feb 19 '14 at 11:25